Solve this following


If the tangent to the curve, $y=f(x)=x \log _{e} x$, $(x>0)$ at a point $(c, f(c))$ is parallel to the line - segement joining the points $(1,0)$ and $(e, e)$, then $\mathrm{c}$ is equal to :

  1. $\frac{1}{e-1}$

  2. $e^{\left(\frac{1}{1-e}\right)}$

  3. $e^{\left(\frac{1}{e-1}\right)}$

  4. $\frac{\mathrm{e}-1}{\mathrm{e}}$

Correct Option: , 3


$f(x)=x \log _{e} x$

$\left.f^{\prime}(\mathrm{X})\right|_{(\mathrm{c}, f(\mathrm{c}))}=\frac{\mathrm{e}-0}{\mathrm{e}-1}$

$f^{\prime}(\mathrm{x})=1+\log _{\mathrm{e}} \mathrm{x}$

$\left.f^{\prime}(\mathrm{x})\right|_{(\mathrm{c}, f(\mathrm{c}))}=1+\log _{\mathrm{e}} \mathrm{c}=\frac{\mathrm{e}}{\mathrm{e}-1}$

$\log _{e} c=\frac{e-(e-1)}{e-1}=\frac{1}{e-1} \Rightarrow c=e^{\frac{1}{e-1}}$

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